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anonymous

2 years ago

\(\{v_1,v_2,\cdots,v_n\}\) are linearly independent means that \(c_1v_1+c_2v_2+\dots+c_nv_n=0\) only for \(c_i=0\). if \(v_2,v_3\dots,v_n\) span \(V\) then we can surely express \(v_1\) in terms of the others, so: $$v_1=k_1v_2+k_2v_3+\dots+k_{n-1}v_n$$but this suggests that we have $$v_1-k_1v_2-k_2v_3-\dots-k_{n-1}v_n=0$$which implies \(v_1,v_2,\dots,v_n\) are linearly dependent -- contradiction

anonymous

2 years ago

since the bottom line suggests we could pick \(c_1=1,c_2=-k_1,c_3=-k_2,\dots,c_n=-k_{n-1}\) to satisfy \(c_1v_1+c_2v_2+\dots+c_nv_n=0\)